Bài 5:

\(m_{Y}=m_{SO_2}+m_{CH_4}=\dfrac{3,36}{22,4}.64+\dfrac{13,44}{22,4}.16=19,2(g)\)

Bài 6:

\(V_{CO_2}=0,15.22,4=3,36(l)\\ V_{NO_2}=0,2.22,4=4,48(l)\\ V_{SO_2}=0,02.22,4=0,448(l)\\ V_{N_2}=0,03.22,4=0,672(l)\)

\(a.V_{NH_3}=0,8.22,4=17,92\left(l\right)\\ b.n_{N_2}=\dfrac{140}{28}=5\left(mol\right)\\ V_{N_2}=112\left(l\right)\\ n_{O_2}=\dfrac{96}{32}=3\left(mol\right)\\ V_{O_2}=22,4.3=67,2\left(l\right)\)

\(c.n_{hh}=0,5+0,2=0,7\left(mol\right)\\ V_{hh}=0,7.22,4=15,68\left(l\right)\\ d.m_{NH_3}=4,25\left(g\right)\\ n_{CO}=\dfrac{28}{22,4}=1,25\left(mol\right)\\ m_{CO}=1,25.28=35\left(g\right)\\ m_{hh}=4,25+35=39,25\left(g\right)\)

a. Thể tích của 0,8 mol khí NH3 (đktc)

=>V_NH3=0,8.22,4=17,92l

b. Thể tích ở đktc của 140 g khí Nitơ ; 96g khí Oxi

n_N2=140\14.2=5 mol

n_O2=96\32=3 mol

=>V_hh=(5+3).22,4=179,2l

c. Thể tích của hỗn hợp khí gồm 0,5 mol Cl2 và 0,2 mol N2 ở điều kiện tiêu chuẩn.

=>V_hh=(0,5+0,2).22,4=15,68l

d. Khối lượng của hỗn hợp 0,25 mol khí NH3 , 28lit khí CO.

n _CO=28\22,4=1,25 mol

=>m_hh=0,25.17+1,25.28=39,25g

\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

\(V_{N_2}=0,1.22,4=2,24\left(l\right)\)

\(V_{CO_2}=0,45.22,4=10,08\left(l\right)\)

\(V_{hh}=5,6+2,24+10,08=17,92\left(l\right)\)

$V_{hh}=22,4(0,25+0,1+0,45)=17,92(lít)$

1. \(n_{O_2}=\frac{V}{22,4}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

2.

\(n_{CO_2}=\frac{m}{M}=\frac{4,4}{44}=0,1\left(mol\right)\)

\(n_{O_2}=\frac{m}{M}=\frac{3,2}{32}=0,1\left(mol\right)\)

\(V_{HC}=n.22,4=\left(0,1+0,1\right).22,4=4,48\left(l\right)\)

1. n= 0.3 mol

Câu 1: \(m_{hh}=0,2\cdot27+0,4\cdot39=21\left(g\right)\)

Câu 2: \(V_{khí}=\left(0,1+0,15\right)\cdot22,4=5,6\left(l\right)\)

Câu 3: \(n_{NO_2}=\dfrac{6,9}{46}=0,15\left(mol\right)\) \(\Rightarrow V_{khí}=\left(0,15+0,15\right)\cdot22,4=6,72\left(l\right)\)

Câu 4: 

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow m_{hh}=0,03\cdot44+0,4\cdot2=2,12\left(l\right)\)

\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)

\(a.\)

\(m_{hh}=0.12\cdot90+0.15\cdot58=19.5\left(g\right)\)

\(b.\)

\(V_{hh}=\left(0.25+0.1+0.05\right)\cdot22.4=8.96\left(l\right)\)

\(c.\)

\(n_A=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(M_A=23\cdot2=46\left(\dfrac{g}{mol}\right)\)

\(m_A=0.45\cdot46=20.7\left(g\right)\)

\(d.\)

\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

Vì CO₂ : O₂ = 2 : 1 

\(\Rightarrow n_{CO_2}=0.2\left(mol\right),n_{O_2}=0.1\left(mol\right)\)

\(m_{hh}=0.2\cdot44+0.1\cdot32=12\left(g\right)\)

\(\overline{M}=\dfrac{12}{0.3}=40\left(\dfrac{g}{mol}\right)\)

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